如何正确获取用户坐标？

Question

我通过这种方式使MapViewController采用CLLocationManagerDelegate协议：

extension MapViewController:CLLocationManagerDelegate{

    func locationManager(_ manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) {

        let usrLocation:CLLocation = locations[0] as CLLocation

        // readable
        print("lat = \(usrLocation.coordinate.latitude)")
        print("lon = \(usrLocation.coordinate.longitude)")

        self.userLocation = usrLocation

        locationManager.stopUpdatingLocation()

    }

    func locationManager(_ manager: CLLocationManager, didFailWithError error: Error)
    {
        print("Error \(error)")
    }

    func locationManager(_ manager: CLLocationManager, didChangeAuthorization status: CLAuthorizationStatus) {

        switch status {
        case .restricted, .denied, .notDetermined:

            // no issue here because I make a very generic request to a server
            // AND I don't need user coordinates

            break

        case .authorizedWhenInUse, .authorizedAlways:

            // here I need user coordinates because I need to request content to a server
            // based on user coordinates but when I check for userLocation
            // at first I find <+0.00000000,+0.00000000> +/- 0.00m (speed -1.00 mps / course -1.00

            print("----------------")
            print(userLocation)
            print("----------------")

            break
        }
    }

我需要.authorizedWhenInUse/.autorizeAlways 案例正文中的用户位置数据，因为我需要发送当前 用户坐标到服务器。问题是当我需要它们时，我首先得到了一个像这样的用户位置：

 <+0.00000000,+0.00000000> +/- 0.00m (speed -1.00 mps / course -1.00>

userLocation 是viewcontroller 类的成员，其值设置在locationManager(_ manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) 中。我应该将请求移至此功能吗？以及如何在这种情况下检查用户状态？我只想向授权该应用程序检索其职位的用户提出请求。我希望解决方案不是计时器和多次检查，但感谢您提供任何帮助:-)

Answer 1

仅仅因为您收到了.authorizedWhenInUse 或.authorizedAlways 并不意味着您已经启动了定位服务。这只是意味着它已被授权。您不应该尝试在 didChangeAuthorization 中检索用户位置。

因此，您应该调用startUpdatingLocation 并将用于将此位置发送到服务器的代码移动到您的didUpdateLocations 方法。